anonymous 4 years ago PLEASE HELP!!!!!!!! what is log base 3 six square root 243 (where 6 is inside the "v" of the square root"

1. anonymous

$\log _{3} \sqrt[6]{243}$

2. anonymous

$\log_{3}\sqrt[6]{243}$ like this?

3. anonymous

yes.

4. anonymous

alright lemme see what i can do

5. anonymous

okay, can u help me with a few others as well? i am really not understanding logs :(

6. anonymous

no problem. You know the rule of logs? Where logANS / logBASE = expo?

7. anonymous

?

8. anonymous

so in this case log [\sqrt[6]{243}\] / log3 = 5/6

9. anonymous

i don't really get it..can you draw it out?

10. anonymous

|dw:1327550139579:dw|

11. anonymous

iam not allowed to use a calculator for these so i have to get it in the exponent form

12. anonymous

$\frac{1}{6}\log_3(243)$ is a start

13. anonymous

and since $243=3^5$ you get $\frac{1}{6}\log_3(3^5)=\frac{5}{6}$

14. anonymous

|dw:1327550493647:dw|

15. anonymous

was i on the right rack?

16. anonymous

Yes. Now use logs to make the expo turn into a coefficient

17. anonymous

how do i do that ...?

18. anonymous

log3^y = log243^(1/6)

19. anonymous

Now you can put the y in front of the log as well as the 1/6 so it becomes ylog3 = (1/6)[log243]

20. anonymous

i am so confused :'''(

21. anonymous

shall we go off satelite73's method? We keep it as a log form.

22. anonymous

i have to do this non calc

23. anonymous

So from the $\sqrt[6]{}$ we can turn it into an exponent as 1/6 so$\log_{3}243^{1/6}$

24. anonymous

i get that now, so im using satelites method as that does not require a calc.

25. anonymous

really your life would be a lot easier if you put the 1/6 out front

26. anonymous

i said that too

27. anonymous

$\log(x^n)=n\log(x)$ so $\log(\sqrt[6]{x})=\frac{1}{6}\log(x)$

28. anonymous

i am really not understanding anything. i just can't do this :(

29. anonymous

The reason we use logrithims for unknown exponential equations is to turn it from an exponent to a coefficient. It then becomes an algebraic equation. So in this situation we have $\log_{3}\sqrt[6]{243}$ We turn the $\sqrt[6]{243}$ into $243^{1/6}$ So right now we have $\log_{3} 243^{1/6}$ now since we have the log, we are allowed to drop the 1/6 down from the exponent level to become the leading coefficient thus making the equation $1/6\log_{3}243$ From here we can change the number 243 into a number with a base of 3. We know that 3^5 = 243 therefore we can write $1/6\log_{3}(3^5)$ And knowing that when a base and the number above it are the same number u can cancel out the log and it's base, leaving you with just the 5. So you are to be left with 1/6 * 5 this is known to be 5/6

30. anonymous

dang. thank you so much! alb to help me with anymore?

31. anonymous